leetcode 202. Happy Number

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Write an algorithm to determine if a number is "happy".

A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.

Example: 19 is a happy number

• 12 + 92 = 82
• 82 + 22 = 68
• 62 + 82 = 100
• 12 + 02 + 02 = 1

很自然的解法是：

class Solution(object):    def isHappy(self, n):        """        :type n: int        :rtype: bool        """                def sum2(num):            ans = 0            while num != 0:                ans += (num % 10)**2                num = num/10            return ans                m = set()        while n not in m:            if n == 1:                return True            m.add(n)            n = sum2(n)        return False

 還有可以使用環路鏈表的判斷方法，一個快慢指標一起賽跑，直到追上：

class Solution(object):    def isHappy(self, n):        """        :type n: int        :rtype: bool        """                def sum2(num):            ans = 0            while num != 0:                ans += (num % 10)**2                num = num/10            return ans                slow = n        fast = sum2(sum2(n))        while slow != fast:            slow = sum2(slow)            fast = sum2(sum2(fast))                        return slow == 1

 java代碼其實非常優雅：

int digitSquareSum(int n) {    int sum = 0, tmp;    while (n) {        tmp = n % 10;        sum += tmp * tmp;        n /= 10;    }    return sum;}bool isHappy(int n) {    int slow, fast;    slow = fast = n;    do {        slow = digitSquareSum(slow);        fast = digitSquareSum(fast);        fast = digitSquareSum(fast);    } while(slow != fast);    if (slow == 1) return 1;    else return 0;}

 數學解法：

class Solution {public:int loop[8] = {4,16,37,58,89,145,42,20};bool inLoop(int n){    for(auto x: loop){        if(x == n) return true;    }    return false;}bool isHappy(int n) {    if(n == 1) return true;    if(inLoop(n)) return false;    int next = 0;    while(n){        next += (n%10)*(n%10);        n /= 10;    }    return isHappy(next);}};

 

proof:1.loop number is less than 162.Assume f(x) is the sum of the squares of x’s digits. let’s say 0 < x <= 9,999,999,999 which is larger than the range of an int. f(x) <= 9^2 * 10 = 810. So no mater how big x is, after one step of f(x), it will be less than 810.The most large f(x) value (x < 810) is f(799) = 211. And do this several times: f(211) < f(199) = 163. f(163) < f(99) = 162. So no mater which x you choose after several times of f(x),it finally fall in the range of [1,162] and can never get out.2.I check every unhappy number in range of [1,162] , they all fall in loop {4,16,37,58,89,145,42,20} ,which means every unhappy number fall in this loop.

其實通過枚舉就應該知道上述<=810範圍內的數字都會落在特定範圍，通過不斷縮小範圍來找規律。=》

Using fact all numbers in [2, 6] are not happy (and all not happy numbers end on a cycle that hits this interval):

bool isHappy(int n) {    while(n>6){        int next = 0;        while(n){next+=(n%10)*(n%10); n/=10;}        n = next;    }    return n==1;}

class Solution(object):    def isHappy(self, n):        """        :type n: int        :rtype: bool        """                def sum2(num):            ans = 0            while num != 0:                ans += (num % 10)**2                num = num/10            return ans                while n > 6:            n = sum2(n)        return n == 1

 

 

leetcode 202. Happy Number